At this point there are a large number of very simple results we can deduce about these operations from the axioms. Some of these follow, and some of them have proofs. The remaining proofs should be considered exercises in manipulating axioms. The aim of these results is to allow us to perform any manipulations which we think are "obviously true" due to our experience with working with numbers. Unless otherwise quantified, the following should hold for all x,y∈R{\displaystyle x,y\in \mathbb {R} }.

Both additive and multiplicative inverses are unique. More formally: If both x+y=0{\displaystyle x+y=0} and x+z=0{\displaystyle x+z=0} then y=z{\displaystyle y=z}; and if both xy=1{\displaystyle xy=1} and xz=1{\displaystyle xz=1} then y=z{\displaystyle y=z} (so that the notations −x{\displaystyle -x} and x−1{\displaystyle x^{-1}} make sense).

Proof: For the case of addition: We have x+y=0{\displaystyle x+y=0} and x+z=0{\displaystyle x+z=0}, so adding y{\displaystyle y} to the latter equation, we get (x+z)+y=0+y{\displaystyle (x+z)+y=0+y}, but then by commutativity and associativity of addition we deduce that (x+y)+z=0+y{\displaystyle (x+y)+z=0+y}, and by our other assumption 0+z=0+y{\displaystyle 0+z=0+y}, and then by identity of addition z=y{\displaystyle z=y}. ◻{\displaystyle \Box }

Proof: First we consider the implication ⟹{\displaystyle \implies }. Suppose x>y{\displaystyle x>y}. By definition, this means that x≠y{\displaystyle x\not =y} and y<x{\displaystyle y<x}. If it were also true that x≤y{\displaystyle x\leq y} then by anti-symmetry we have x=y{\displaystyle x=y}, which is impossible. Thus ¬x≤y{\displaystyle \neg x\leq y}.

Conversely, suppose ¬x≤y{\displaystyle \neg x\leq y}. First, if we had x=y{\displaystyle x=y} then by reflexivity x≤y{\displaystyle x\leq y}, which is impossible, so in fact x≠y{\displaystyle x\not =y}. Secondly, by totality we deduce that y≤x{\displaystyle y\leq x}. These two conditions are exactly those required for x>y{\displaystyle x>y}. ◻{\displaystyle \Box }

x<y⟺¬x≥y{\displaystyle x<y\iff \neg x\geq y}

x{\displaystyle x} is non-positive if and only if x{\displaystyle x} is not positive

x{\displaystyle x} is non-negative if and only if x{\displaystyle x} is not negative

If x{\displaystyle x} is both non-positive and non-negative then x=0{\displaystyle x=0}

x{\displaystyle x} is not both positive and negative

x≥0⟺−x≤0{\displaystyle x\geq 0\iff -x\leq 0}

Proof: Suppose x≥0{\displaystyle x\geq 0}. By one of the axioms we get x+(−x)≥0+(−x){\displaystyle x+(-x)\geq 0+(-x)}. By additive inverse this gives 0≥0+(−x){\displaystyle 0\geq 0+(-x)} and then by additive identity 0≥−x{\displaystyle 0\geq -x}, as required.

Proof: By totality of the order, we have either x≥0{\displaystyle x\geq 0} or x≤0{\displaystyle x\leq 0}. In the first case we can apply the axiom linking the order to multiplication directly to 0≤x{\displaystyle 0\leq x} and deduce 0≤x2{\displaystyle 0\leq x^{2}}. In the latter case we apply the last result in this list to 0≤x{\displaystyle 0\leq x} and obtain x2≥0{\displaystyle x^{2}\geq 0}. ◻{\displaystyle \Box }

Although it might be said that the entirety of this book is devoted to studying the applications of completeness, there are in particular some simple applications we can give easily which provide an indication as to how completeness solves the problem with the rationals described above.

We deal only with the case x≥1{\displaystyle x\geq 1}. The case x∈[0,1){\displaystyle x\in [0,1)} is left for the exercises.

First we note that when y,z∈R{\displaystyle y,z\in \mathbb {R} } are non-negative, y<z⟹y2<z2{\displaystyle y<z\implies y^{2}<z^{2}} (In the terminology we will introduce later, this says that the function y↦y2{\displaystyle y\mapsto y^{2}} is strictly increasing). This makes it clear that there can be only one square root of x{\displaystyle x}, and so it remains to find one.

Let S={y∈R:y2≤x}{\displaystyle S=\{y\in \mathbb {R} :y^{2}\leq x\}}. We wish to apply the least upper bound axiom to S{\displaystyle S}, so we must show that it is non-empty and bounded above.

That S{\displaystyle S} is non-empty is clear, since 1∈S{\displaystyle 1\in S}.

Furthermore, x{\displaystyle x} itself is an upper bound for S{\displaystyle S}, since if y>x≥1{\displaystyle y>x\geq 1}, then y2>y{\displaystyle y^{2}>y}, so that y2>x{\displaystyle y^{2}>x}, and hence y∉S{\displaystyle y\not \in S}.

Putting these facts together, by the least upper bound axiom, we deduce that S{\displaystyle S} has a least upper bound, which we call s{\displaystyle s}. We wish to show that s{\displaystyle s} is the square root of x{\displaystyle x} that we seek.

Certainly s{\displaystyle s} is positive, since 1∈S{\displaystyle 1\in S} and so s≥1{\displaystyle s\geq 1}. In particular, we may divide by s{\displaystyle s}.

To show that s2=x{\displaystyle s^{2}=x}, we eliminate the possibilities that s2>x{\displaystyle s^{2}>x}, and that s2<x{\displaystyle s^{2}<x}.

So t{\displaystyle t} is in fact an upper bound for S{\displaystyle S}, but this is impossible, since t<s{\displaystyle t<s} and s{\displaystyle s} is the least upper bound for S{\displaystyle S}.

Thus we have concluded that s2≤x{\displaystyle s^{2}\leq x}.

Now suppose that s2<x{\displaystyle s^{2}<x}. Let t=s+x−s22s{\displaystyle t=s+{\frac {x-s^{2}}{2s}}}. In a similar manner to the above, we deduce that t2<x{\displaystyle t^{2}<x}, so t∈S{\displaystyle t\in S}, but this is impossible since t>s{\displaystyle t>s} and s{\displaystyle s} is an upper bound for S{\displaystyle S}.

Thus we have concluded that s2≥x{\displaystyle s^{2}\geq x}, and so s2=x{\displaystyle s^{2}=x} as required.◻{\displaystyle \Box }

This argument may appear excessively complex (especially since some details are left for the exercises), and indeed there is a sense in which it is, and we shall be able to present a much neater argument later. Nevertheless, it suffices to show that we can find a square root of 2, and so avoid the immediate problem with the rationals posed at the beginning of this section. To show that no more elaborate construction will give rise to the same problem will have to wait until we reach the study of continuity.

but this is precisely the statement that N{\displaystyle \mathbb {N} } is bounded above. Certainly also it is non-empty, so we can apply the completeness axiom to get a least upper bound for N{\displaystyle \mathbb {N} }. Call this least upper bound l{\displaystyle l}.

Since l{\displaystyle l} is a least upper bound, we know that l−1{\displaystyle l-1} is not an upper bound, and thus ∃n∈N:n>l−1{\displaystyle \exists n\in \mathbb {N} :n>l-1}. But then, n+1>l{\displaystyle n+1>l}, and n+1∈N{\displaystyle n+1\in \mathbb {N} } so we get the contradiction that l{\displaystyle l} is not an upper bound for N{\displaystyle \mathbb {N} } after all.

Thus, our supposition was false, and (a) holds.

b) Take x∈R+{\displaystyle x\in \mathbb {R} ^{+}}. Certainly x≠0{\displaystyle x\not =0}, so that we can invert x{\displaystyle x} to get x−1∈R+{\displaystyle x^{-1}\in \mathbb {R} ^{+}}. Applying part (a) to x−1{\displaystyle x^{-1}}, we can find n∈N{\displaystyle n\in \mathbb {N} } with n>x−1{\displaystyle n>x^{-1}}, and then inverting this inequality, we deduce 1n<x{\displaystyle {\frac {1}{n}}<x} as required.◻{\displaystyle \Box }

Putting this together with the fact that xn<yn−1{\displaystyle xn<yn-1} deduced above, we get:

N−m>xn{\displaystyle N-m>xn}

So, in summary, we have yn>N−m>xn{\displaystyle yn>N-m>xn}, so y>N−mn>x{\displaystyle y>{\frac {N-m}{n}}>x}, and we have found the rational number we want.

To find an irrational number, we use what we have just deduced to first find a rational q∈(x+2,y+2){\displaystyle q\in (x+{\sqrt {2}},y+{\sqrt {2}})}, so that q−2∈(x,y){\displaystyle q-{\sqrt {2}}\in (x,y)}. Furthermore, q−2{\displaystyle q-{\sqrt {2}}} must be irrational, for if it were rational then we would also have q−(q−2)=2{\displaystyle q-(q-{\sqrt {2}})={\sqrt {2}}} rational, and we know that it is not. ◻{\displaystyle \Box }

We often need to take a sum or product of several real numbers at a time. Since "..." is given no meaning by our axioms, we can't just write "a1+a2+⋯+an{\displaystyle a_{1}+a_{2}+\dots +a_{n}}". Thus we use the symbols ∑k=1nak{\displaystyle \sum _{k=1}^{n}a_{k}} and ∏k=1nak{\displaystyle \prod _{k=1}^{n}a_{k}} to denote the sum and product, respectively, over an arbitrary finite number of real numbers. We do this inductively, as follows:

The order of summation can be changed arbitrarily. That is, if {ak:1≤k≤n}={bk:1≤k≤n}{\displaystyle \{a_{k}:1\leq k\leq n\}=\{b_{k}:1\leq k\leq n\}}, then ∑k=1nak=∑k=1nbk{\displaystyle \sum _{k=1}^{n}a_{k}=\sum _{k=1}^{n}b_{k}} and ∏k=1nak=∏k=1nbk{\displaystyle \prod _{k=1}^{n}a_{k}=\prod _{k=1}^{n}b_{k}}